NIMCET 2021 Mathematics PYQ — The eccentric angle of the extremities of latusrectum of the elli… | Mathem Solvex | Mathem Solvex
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NIMCET 2021 — Mathematics PYQ
NIMCET | Mathematics | 2021
The eccentric angle of the extremities of latusrectum of the ellipse a2x2+b2y2=1 are given by
Choose the correct answer:
A.
tan−1(±bae)
B.
tan−1(±ebe)
C.
tan−1(±aeb)
(Correct Answer)
D.
tan−1(±bea)
Correct Answer:
tan−1(±aeb)
Explanation
1. Coordinates of the Latus Rectum Extremities
The latus rectum of an ellipse is a chord passing through the focus (ae,0) and perpendicular to the major axis. The coordinates of the endpoints (extremities) in the first quadrant and fourth quadrant are:
(x,y)=(ae,±ab2)
2. Parametric Coordinates and Eccentric Angle
Any point on the ellipse a2x2+b2y2=1 can be represented in parametric form as:
x=acosϕ
y=bsinϕ
Where ϕ is the eccentric angle.
3. Equating the Coordinates
To find the eccentric angle ϕ for the latus rectum, we equate the parametric forms to the coordinates found in Step 1:
acosϕ=ae⟹cosϕ=e
bsinϕ=±ab2⟹sinϕ=±ab
4. Finding the Tangent of the Angle
To find ϕ in terms of tan−1, we divide sinϕ by cosϕ:
tanϕ=cosϕsinϕ=e±b/a=±aeb
Therefore:
ϕ=tan−1(±aeb)
5. Simplifying using Eccentricity (e)
We know that for an ellipse, b2=a2(1−e2), which means ab=1−e2.
Substituting this into our tanϕ expression:
tanϕ=e±1−e2
ϕ=tan−1(e±1−e2)
Final Answer
Comparing our result with the given options, we find that the expression matches Option (D).
Correct Option: (D) tan−1(e±1−e2)
Explanation
1. Coordinates of the Latus Rectum Extremities
The latus rectum of an ellipse is a chord passing through the focus (ae,0) and perpendicular to the major axis. The coordinates of the endpoints (extremities) in the first quadrant and fourth quadrant are:
(x,y)=(ae,±ab2)
2. Parametric Coordinates and Eccentric Angle
Any point on the ellipse a2x2+b2y2=1 can be represented in parametric form as:
x=acosϕ
y=bsinϕ
Where ϕ is the eccentric angle.
3. Equating the Coordinates
To find the eccentric angle ϕ for the latus rectum, we equate the parametric forms to the coordinates found in Step 1:
acosϕ=ae⟹cosϕ=e
bsinϕ=±ab2⟹sinϕ=±ab
4. Finding the Tangent of the Angle
To find ϕ in terms of tan−1, we divide sinϕ by cosϕ:
tanϕ=cosϕsinϕ=e±b/a=±aeb
Therefore:
ϕ=tan−1(±aeb)
5. Simplifying using Eccentricity (e)
We know that for an ellipse, b2=a2(1−e2), which means ab=1−e2.
Substituting this into our tanϕ expression:
tanϕ=e±1−e2
ϕ=tan−1(e±1−e2)
Final Answer
Comparing our result with the given options, we find that the expression matches Option (D).